3
\$\begingroup\$

I'm trying to build my own relay board to use with a 3.3V controller. I found the schematic of the board, so I'll try to replicate it. I'm only having trouble understanding what's the purpose of the LL4148 diode parallel to the coils.

Here is the schematic: enter image description here

Here is a simulation of the circuit. Removing the diode seens to make no difference.

http://bit.do/fgPsz

\$\endgroup\$
  • 1
    \$\begingroup\$ youtube.com/watch?v=LXGtE3X2k7Y \$\endgroup\$ – G36 Nov 11 at 14:35
  • 9
    \$\begingroup\$ Drawing schematics upside down is non standard. You should flip it around. \$\endgroup\$ – DKNguyen Nov 11 at 14:38
  • 6
    \$\begingroup\$ Without the diode, you will damage Q1. \$\endgroup\$ – Bimpelrekkie Nov 11 at 14:39
  • 2
    \$\begingroup\$ @DKNguyen: maybe Josy Sclei is from Australia \$\endgroup\$ – Curd Nov 11 at 15:06
9
\$\begingroup\$

From the question below:

A diode is put in parallel with a relay coil (with opposite polarity) to prevent damage to other components when the relay is turned off.

This is called a flyback diode.

See also How to choose a flyback diode for a relay?

\$\endgroup\$
  • 1
    \$\begingroup\$ Right, to prevent current spikes when the coils is latched. Now I remember seeing that in class, thank you very much! \$\endgroup\$ – Josy Sclei Nov 11 at 14:37
  • 5
    \$\begingroup\$ @JosySclei - it's actually to prevent "Voltage spikes", not current spikes. And the problem is during turn-off, not turn-on. At turn-off when current is already flowing in the relay inductance the current will try to keep flowing even when the transistor switches off. The diode provides a path for the current to flow. \$\endgroup\$ – Kevin White Nov 11 at 22:07
4
\$\begingroup\$

The diode is known as freewheeling/flyback diode.

Imagine that the relay is energised. A current will flow through it and will set up a magnetic field around it. When the relay is de-energised, this field will collapse as there is no more current to sustain that field. Michael Faraday found out that whenever a magnetic field associated with a coil changes, a voltage will be induced in it. In our relay, this change in field is rapid and the voltage will be high. This can be damaging to the transistor. So we put a diode across the coil such that it will short circuit the coil and the voltage across it quickly settles to zero.

I believe this will help.

\$\endgroup\$
  • 1
    \$\begingroup\$ What happens is that the current flowing through the relay will commute from the transistor to the diode. Because the relay current is not abruptly cut off, the high voltage spike will not occur. The reverse voltage will not be higher than the forward voltage of the diode, with the current equal to the initial relay current. On a side note, because the relay current is not discontinous, the magnetic field will also not abruptly change. The relay will therefor switch off more slowly, which may cause micro-welding of the contacts. \$\endgroup\$ – Bart Nov 11 at 15:10
  • \$\begingroup\$ Pretty much what I said.The reason relay current won't change abruptly is self induction of relay coil and the freewheeling diode. \$\endgroup\$ – ASWIN VENU Nov 11 at 15:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.